HDU 5569 matrix

题目传送门

中文翻译:

解题思路:

f[i][j]表示到第i行第j列时,可获得的最小收益

AC代码:

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

long long n,m,a[1001][1001],f[1001][1001]; 

int main() {
    while(scanf("%d%d",&n,&m) != EOF) {
        memset(a,0,sizeof(a));
        memset(f,0x3f3f3f,sizeof(f));
        for(int i = 1;i <= n; i++)
            for(int j = 1;j <= m; j++) 
                scanf("%d",&a[i][j]);
        f[1][1] = a[1][1];
        for(int i = 2;i <= m; i++) {//处理第一行 
            if(i % 2 == 0) f[1][i] = f[1][i-1] + a[1][i] * a[1][i-1];
            else f[1][i] = f[1][i-1];
            if(i == 2) f[1][2] -= f[1][1];
        }
        for(int i = 2;i <= n; i++) {//处理第一列 
            if(i % 2 == 0) f[i][1] = f[i-1][1] + a[i][1] * a[i-1][1];
            else f[i][1] = f[i-1][1];
            if(i == 2) f[2][1] -= f[1][1];
        }
        for(int i = 2;i <= n; i++)
            for(int j = 2;j <= m; j++) {
                if((j + i) % 2 == 0)//走了奇数步 
                    f[i][j] = min(f[i-1][j],f[i][j-1]);
                else //走了偶数步 
                    f[i][j] = min(f[i-1][j] + a[i-1][j] * a[i][j],f[i][j-1] + a[i][j-1] * a[i][j]);
            }
        printf("%d
",f[n][m]);
    }
    return 0;
}